Write as a Set of Linear Factors 6x^3+37x^2-34x+7=0

$6 x^{3} + 37 x^{2} - 34 x + 7 = 0$

Step 1

Factor $6 x^{3} + 37 x^{2} - 34 x + 7$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 7$

$q = \pm 1, \pm 6, \pm 2, \pm 3 = 0$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 7, \pm 1.1\overline{6}, \pm 3.5, \pm 2.\overline{3} = 0$

Substitute $0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $0.5$ is a root of the polynomial.

Substitute $0.5$ into the polynomial.

$6 \cdot {0.5}^{3} + 37 \cdot {0.5}^{2} - 34 \cdot 0.5 + 7 = 0$

Raise $0.5$ to the power of $3$ .

$6 \cdot 0.125 + 37 \cdot {0.5}^{2} - 34 \cdot 0.5 + 7 = 0$

Multiply $6$ by $0.125$ .

$0.75 + 37 \cdot {0.5}^{2} - 34 \cdot 0.5 + 7 = 0$

Raise $0.5$ to the power of $2$ .

$0.75 + 37 \cdot 0.25 - 34 \cdot 0.5 + 7 = 0$

Multiply $37$ by $0.25$ .

$0.75 + 9.25 - 34 \cdot 0.5 + 7 = 0$

Add $0.75$ and $9.25$ .

$10 - 34 \cdot 0.5 + 7 = 0$

Multiply $- 34$ by $0.5$ .

$10 - 17 + 7 = 0$

Subtract $17$ from $10$ .

$- 7 + 7 = 0$

Add $- 7$ and $7$ .

$0 = 0$

Since $0.5$ is a known root, divide the polynomial by $2 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} + 37 x^{2} - 34 x + 7}{2 x - 1} = 0$

Divide $6 x^{3} + 37 x^{2} - 34 x + 7$ by $2 x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$= 0$

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $2 x$ .

$= 0$

Multiply the new quotient term by the divisor.

$= 0$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x^{3} - 3 x^{2}$

$= 0$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$= 0$

Pull the next terms from the original dividend down into the current dividend.

$= 0$

Divide the highest order term in the dividend $40 x^{2}$ by the highest order term in divisor $2 x$ .

$= 0$

Multiply the new quotient term by the divisor.

$= 0$

The expression needs to be subtracted from the dividend, so change all the signs in $40 x^{2} - 20 x$

$= 0$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$= 0$

Pull the next terms from the original dividend down into the current dividend.

$= 0$

Divide the highest order term in the dividend $- 14 x$ by the highest order term in divisor $2 x$ .

$= 0$

Multiply the new quotient term by the divisor.

$= 0$

The expression needs to be subtracted from the dividend, so change all the signs in $- 14 x + 7$

$= 0$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$= 0$

Since the remander is $0$ , the final answer is the quotient.

$3 x^{2} + 20 x - 7 = 0$

Write $6 x^{3} + 37 x^{2} - 34 x + 7$ as a set of factors.

$(2 x - 1) (3 x^{2} + 20 x - 7) = 0$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + i x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 7 = - 21$ and whose sum is $i = 20$ .

Factor $20$ out of $20 x$ .

$(2 x - 1) (3 x^{2} + 20 (x) - 7) = 0$

Rewrite $20$ as $- 1$ plus $21$

$(2 x - 1) (3 x^{2} + (- 1 + 21) x - 7) = 0$

Apply the distributive property.

$(2 x - 1) (3 x^{2} - 1 x + 21 x - 7) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x - 1) ((3 x^{2} - 1 x) + 21 x - 7) = 0$

Factor out the greatest common factor (GCF) from each group.

$(2 x - 1) (x (3 x - 1) + 7 (3 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(2 x - 1) ((3 x - 1) (x + 7)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (3 x - 1) (x + 7) = 0$

Write as a Set of Linear Factors 6x^3+37x^2-34x+7=0

Related Questions

Support LCMGCF.com by disabling your adblocker.